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Light geodesics in FLWR

  1. Jan 12, 2010 #1
    hello,

    I developed an application that models the trajectory of a light geodesic in the FLWR metric leaving from a galaxy and coming to our. I made for the moment the euclidean case (k=0) with a zero cosmological constant . So, the metric can be written :

    [tex]ds^{2}=c^{2}dt^{2}-R(t)^{2}d\sigma^{2} [/tex]​

    with
    [tex]
    d\sigma^{2}=dr^2+r^2(d\theta^2+sin^2\theta\,d\phi^2)
    [/tex]​

    and
    [tex]R(t)=R_{0}\,(\frac{3\,H_{0}}{2}\,t)^{2/3} [/tex]​

    I get differential system from geodesics general equation :

    [tex]\frac{d^2\,u^{i}}{ds^2}+\Gamma_{j}_{}^{i}_{k}\,\frac{d\,u^j}{ds}\,\frac{d\,u^k}{ds}=0[/tex]​

    here is the differential system :

    [tex]
    \frac{d^{2}ct}{ds^{2}} = -\frac{2}{3}\,R_{0}^{2}\,\bigg(\frac{3\,H_{0}}{2\,c}\bigg)^{4/3}\,(ct)^{1/3}\,\bigg(\bigg(\frac{d\,r}{ds}\bigg)^{2}+r^2\,\bigg(\frac{d\,\theta}{ds}\bigg)^{2}+r^{2}\,sin^{2}(\theta)\,\bigg(\frac{d\,\phi}{ds}\bigg)^{2}\bigg)[/tex]

    [tex]\frac{d^{2}r}{ds^{2}} & = & -\frac{4}{3\,ct}\,\frac{d\,ct}{ds}\,\frac{d\,r}{ds}+r\,\bigg(\bigg(\frac{d\,\theta}{ds}\bigg)^{2}-sin^{2}(\theta)\,\bigg(\frac{d\,\phi}{ds}\bigg)^{2}\bigg)[/tex]
    [tex]\frac{d^{2}\theta}{ds^{2}} & = & -\frac{2}{r}\,\frac{d\,r}{ds}\,\frac{d\,\theta}{ds}-\frac{4}{3\,ct}\,\frac{d\,ct}{ds}\,\frac{d\,\theta}{ds}+\frac{sin(2\,\theta)}{2}\,\bigg(\frac{d\,\phi}{ds}\bigg)^{2} [/tex]

    [tex]\frac{d^{2}\phi}{ds^{2}} & = & -2\,\frac{d\,\phi}{ds}\,\bigg(\frac{2}{3\,ct}\,\frac{d\,ct}{ds}+cotan(\theta)\,\frac{d\,\theta}{ds}+\frac{1}{r}\,\frac{d\,r}{ds}\bigg)[/tex]​

    I solve this system by Runge-Kutta numerical method. the initial conditions are linked by the following equation :

    [tex]\bigg(\frac{d\,ct}{ds}\bigg)_{0}^{2}=R(t_{0})^{2}\bigg(\bigg(\frac{dr}{ds}\bigg)_{0}^{2}+r_{0}^2\bigg(\bigg(\frac{d\theta}{ds}\bigg)_{0}^{2}+sin^{2}(\theta_{0})\bigg(\frac{d\phi}{ds}\bigg)_{0}^{2}\bigg)\bigg)[/tex]​

    If i take [tex] \frac{d\,\phi}{ds} = \frac{d\,\theta}{ds} = 0 [/tex], ie if i consider only radial trajectories, i get the figure 1 in attachment where i plot the Distance "r" as a function of Local time "ct" expressed in Megaparsec. i took to get this figure :

    [tex]\bigg(\frac{d\,r}{ds}\bigg)_{0} = -1 [/tex]

    [tex]\bigg(\frac{d\,\theta}{ds}\bigg)_{0} = 0 [/tex]
    [tex] \bigg(\frac{d\,\phi}{ds}\bigg)_{0} = 0 [/tex]
    [tex] r{0} = 3000 Mpc [/tex]
    and
    [tex]R_{0} = 1 [/tex]​

    The dotted line represents the trajectory of light ray and the solid line represents the trajectory of the galaxy where light is emitted.
    the problem I encounter is when I'm looking for the propagation of two rays and to compare the actual angular size of the object and the angular size that we observe. This is represented on figure 2 in attachment. I don't know how to make converge the two rays when they arrive in our galaxy. I think that "r" and "theta" should vary simultaneously but what initial conditions should I take to it ?

    We can see on figure 3 in attachment the case where curvature k=0. It's possible to have "theta" constant and "r" which decreases but in the two others cases, "theta" and "r" vary in same time. I hope you understand my problem.

    Thanks.
     

    Attached Files:

    Last edited: Jan 12, 2010
  2. jcsd
  3. Jan 13, 2010 #2

    Wallace

    User Avatar
    Science Advisor

    The way that this is sometimes addressed in general is via the Dyer-Roder equation, which looks at a bundle of light rays and works out how that bundle get converged or sheared as it passes through the Universe. However in the case of a smooth universe (as you are using) this becomes much simpler to compute, and there is a relatively simple single integral to be performed, which tells you this 'Angular Diameter Distance'.

    There is an excellent summary of this and other distance measures available from ArXiv http://arxiv.org/abs/astro-ph/9905116" [Broken]. Every cosmology grad student should have this paper handy at all times (and most that I know do...).
     
    Last edited by a moderator: May 4, 2017
  4. Feb 1, 2010 #3
    Hello,

    i would like to get the curve in attachment ("dda_2.jpg") for the opened/closed/euclidean cases in FLWR metric, with numerical approach. I have seen informations about the definition of the angular diameter distance and my problem lies on the variation of theta ( i only take into account of theta and r). According the definition of angular diameter distance :

    [tex] d_{\theta}=R(t_{emission})\,r_{1} = a(t_{emission})\, R_{0}\,r_{1}=a(t_{emission}))\, d_{comobile,today} [/tex]​

    with
    [tex]a(t)=\frac{R(t)}{R_{0}} [/tex].​

    I solved numerically a(t) thanks to Friedmann equations. I have with the definition of FLWR metric :

    [tex]\int_{r}^{r1}\,\frac{dr}{\sqrt{1-k\,r^2}} = \int_{t1}^{t}\, \frac{c dt}{R(t)}[/tex]​
    where k=-1,0,1

    For example, i can write in euclidean case :

    [tex]r(t)=r1-\int_{t1}^{t}\, \frac{c dt}{R(t)}[/tex]​

    this implies that r(t) is going to decrease with time and so, the angle theta is going to increase indefinetely because r(t) decreases until it vanishes, according this formula ( D is the diameter of the emitting object ) :

    [tex]\theta=\frac{D}{r(t)}[/tex]​
    .

    That's not the behavior of the angle theta on "dda_2.jpg" which doesn't reach an infinite value for r=0.


    In this last expression, have i got to multiply r(t) by R(t) ? , this would give :

    [tex]\theta=\frac{D}{R(t)\,r(t)}[/tex]​
    .

    I think light geodesics are not like comobile object (one galaxy for example) whose distance is (r1 * R(t)) where r1 doesn't vary over time, that's why i use r(t).

    I hope you will understand my problem with the simultaneous variation of r and theta. I repeat, i'am searching to get the figure "dda_2.jpg" with the 3 cases (open/close/euclidean).

    Thanks in advance.
     

    Attached Files:

    Last edited: Feb 1, 2010
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